Question: Graph this system of equations and solve. $2x-3y = 6$ $-9x+3y = 15$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $2x-3y = 6$ , to slope-intercept form. $y = \dfrac{2}{3} x - 2$ The y-intercept for the first equation is $-2$ , so the first line must pass through the point $(0, -2)$ The slope for the first equation is $\dfrac{2}{3}$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move up You must also move $3$ positions to the right. $3$ positions to the right. $2$ positions up from $(0, -2)$ is $(3, 0)$ Graph the blue line so it passes through $(0, -2)$ and $(3, 0)$ Convert the second equation, $-9x+3y = 15$ , to slope-intercept form. $y = 3 x + 5$ The y-intercept for the second equation is $5$ , so the second line must pass through the point $(0, 5)$ The slope for the second equation is $3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up You must also move $1$ positions to the right. $1$ position to the right. $3$ positions up from $(0, 5)$ is $(1, 8)$ Graph the green line so it passes through $(0, 5)$ and $(1, 8)$ The solution is the point where the two lines intersect. The lines intersect at $(-3, -4)$.